Find the derivative of the given function.y=4arccos(3x6−x5)y=4\arccos\left(3x^6-x^5\right)y=4arccos(3x6−x5)

− 4 ( 18 x 5 − 5 x 4 ) 1 − ( 3 x 6 + x 5 ) 2 \frac{-4\left(18x^5-5x^4\right)}{\sqrt{1-\left(3x^6+x^5\right)^2}}

Find the derivative of the given function. y = 4 arccos ( 3 x 6 − x 5 ) y=4\arccos\left(3x^6-x^5\right) y = 4 arccos ( 3 x 6 − x 5 )

Here, we wanna find the derivative of the function y equals 4 times the arccosine of 3x to the power of 6 minus x to the power of 5. Now remember that arccosine is the same thing as inverse cosine written like this. Those mean the same exact thing. So in finding this derivative here, y prime, we have this constant 4 that we can just stick out front and just focus in on this derivative here. Now since I'm taking the arccosine of something that is not just x, that means that I have to apply the chain rule here. And, also, when I'm taking the derivative of this outside function, this inverse cosine, I need to remember to put not just x under that square root, but 3x to the power of 6 minus x to the power of 5. So starting with that outermost function, I know that the derivative of arccosine is negative one over the square root of 1 minus whatever I'm taking the arccosine of squared. So this is 3x to the power of 6 minus x to the power of 5, and this gets squared. So make sure that you include everything there. Then we multiply this using the chain rule by the derivative of this, 3x to the power of 6 minus x to the power of 5. Using our power rule, pulling that 6 out front and multiplying by it, this gives me 18x to the power of 5, decreasing that power by 1, and then minus 5x to the power of 4. Now we can condense this all into one fraction instead of being all split up like this. This will then give me negative 4 times 18x to the power of 5 minus 5x to the power of 4 over this very pretty square root, 1 minus 3x to the 6th plus x to the 5th squared, and this is my full derivative. Sometimes these derivatives might not end up looking very pretty, but we've still done them correctly. Let us know if you have any questions here.

Find the derivative of the given function. y = 4 arccos ⁡ ( 3 x 6 − x 5 ) y=4\arccos\left(3x^6-x^5\right) y = 4 arccos ( 3 x 6 − x 5 )

− 4 ( 18 x 5 − 5 x 4 ) 1 − ( 3 x 6 + x 5 ) 2 \frac{-4\left(18x^5-5x^4\right)}{\sqrt{1-\left(3x^6+x^5\right)^2}}

− 4 1 − ( 3 x 6 − x 5 ) 2 -\frac{4}{\sqrt{1-\left(3x^6-x^5\right)^2}} − 1 − ( 3 x 6 − x 5 ) 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> 4

− 4 ( 18 x 5 − 5 x 4 ) 1 − x 2 -\frac{4\left(18x^5-5x^4\right)}{\sqrt{1-x^2}}

− 4 ( 18 x 5 − 5 x 4 ) sin ⁡ − 1 ( 3 x 6 − x 5 ) -4\left(18x^5-5x^4\right)\sin^{-1}\left(3x^6-x^5\right) − 4 ( 18 x 5 − 5 x 4 ) sin − 1 ( 3 x 6 − x 5 )

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Inverse Trigonometric Functions

Calculus

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Inverse Trigonometric Functions

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Find the derivative of the given function.
$y=4\arccos\left(3x^6-x^5\right)$

$-\frac{4}{\sqrt{1-\left(3x^6-x^5\right)^2}}$

$\frac{- 4 (18 x^{5} - 5 x^{4})}{\sqrt{1 - {(3 x^{6} + x^{5})}^{2}}}$

$- \frac{4 (18 x^{5} - 5 x^{4})}{\sqrt{1 - x^{2}}}$

$-4\left(18x^5-5x^4\right)\sin^{-1}\left(3x^6-x^5\right)$

Verified step by step guidance

Identify the function to differentiate: y = 4 * arccos(3x^6 - x^5). This is a composition of functions, so we will use the chain rule.

Recall the derivative of arccos(u) with respect to u is -1/sqrt(1-u^2). Here, u = 3x^6 - x^5.

Apply the chain rule: dy/dx = dy/du * du/dx. First, find dy/du = -4/sqrt(1-(3x^6-x^5)^2).

Next, find du/dx by differentiating u = 3x^6 - x^5 with respect to x. This gives du/dx = 18x^5 - 5x^4.

Combine the results using the chain rule: dy/dx = (-4/sqrt(1-(3x^6-x^5)^2)) * (18x^5 - 5x^4).

7:26m

Watch next

Master Derivatives of Inverse Sine & Inverse Cosine with a bite sized video explanation from Patrick

Start learning

Derivatives of Inverse Trigonometric Functions practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice